lecture 4 fourier series
TRANSCRIPT
8/13/2019 Lecture 4 fourier series
http://slidepdf.com/reader/full/lecture-4-fourier-series 1/18
1/30/2014
MATH 5311 – Advanced Engineering Math
Fourier Cosine and Sine Transforms
Section 11.8
Fourier Cosine Transform
0
cos f x A xd
0
2cos A f x x dx
Recall the Fourier cosine integral
Substitute A(ω ) into the Fourier cosine integral.
0 0
2cos cos f x f x x dx x d
0 0
2 2cos cos f x x dx x d
ˆ
C f
Fourier CosineTransform
0
2ˆ
cosC C f f x x dx f x
F
Inverse FourierCosine Transform
0
2 ˆ ˆ
cos -1C C C f x f x d f
F
Page534
8/13/2019 Lecture 4 fourier series
http://slidepdf.com/reader/full/lecture-4-fourier-series 2/18
1/30/2014
Fourier Sine Transform
Fourier SineTransform
0
2ˆ
sinS S f f x x dx f x
F
Inverse FourierSine Transform
0
2 ˆ ˆ
sin -1S S S f x f x d f
F
Page 535
Linearity of Fourier transforms
Theorem – Let a and b be constants and f and g be functions withFourier cosine transforms. Then
C C C af x bg x a f x b g x F F F
Proof – The Fourier cosine transform of is af x bg x
0
2cosC af x bg x af x bg x x dx
F
0
2cos cosaf x x bg x x dx
0 0
2 2cos cosaf x x dx bg x x dx
0 0
2 2cos cosa f x x dx b g x x dx
C C a f x b g x F F
8/13/2019 Lecture 4 fourier series
http://slidepdf.com/reader/full/lecture-4-fourier-series 3/18
1/30/2014
Fourier transforms of derivatives
Theorem – Let f be a function with a Fourier sine transform and assume
lim 0 x f x
2' 0C s f x f f x
F F Then
Proof
0
2' ' cosC f x f x x dx
F
cos '
sin
u x dv f x dx
du x dx v f x
0
0
2cos sin f x x f x x dx
0
2 20 sin f f x x dx
20 s f f x
F
Fourier transforms of derivatives
2
2
2' 0
'
2'' ' 0
2'' 0
C s
S C
C C
s S
f x f f x
f x f x
f x f f x
f x f f x
F F
F F
F F
F F
8/13/2019 Lecture 4 fourier series
http://slidepdf.com/reader/full/lecture-4-fourier-series 4/18
1/30/2014
MATH 5311 – Advanced Engineering Math
Fourier Transform, Discrete and FastFourier Transforms
Section 11.9
Deborah Koslover
RBN 4010
Complex form of the Fourier Integral
0
cos sin f x A x B x d 1
sin B f d
1cos A f d
Substitute A and B into the Fourier integral.
Simplify. Note – The sine and cosine can be brought into the integralsbecause they have different variables.
0
1 1cos cos sin sin f x f dx f d x d
Combine inner integrals into one integral.
0
1cos cos sin sin f x f x x d d
( )
( ) x
8/13/2019 Lecture 4 fourier series
http://slidepdf.com/reader/full/lecture-4-fourier-series 5/18
1/30/2014
Complex form of the Fourier Integral
0
1cos cos sin sin f x f x x d d
Use the trigonometric identity cos cos sin sin cos A B A B B A
0
1cos f x f x d d
Notice that cos f x d g is a function of ω .
and further that it is an even function of ω .
cos cos
cos
g f x d f x d
f x d g
So
0
1 12
f x g d g d
1cos
2 f x d d
Complex form of the Fourier Integral
1cos
2 f x f x d d
sinh f x d The new function is an odd function of ω .
sin sin
sin
h f x d f x d
f x d h
So 0 sin
2 2i i
h d f x d d
Therefore,
1cos 0
2 f x f x d d
1cos sin
2 2i
f x f x d d f x d d
8/13/2019 Lecture 4 fourier series
http://slidepdf.com/reader/full/lecture-4-fourier-series 6/18
1/30/2014
Complex form of the Fourier Integral
1cos sin
2 2i
f x f x d d f x d d
Simplify.
1cos sin
2 f x f x d d f i x d d
Combine into one integral.
1cos sin
2 f x f x i x d d
Using Euler’s equation, cos sin ii e , the integral become
1
2i x
f x f e d d
This is the complex Fourier integral .
Fourier Transform
12
i x f x f e d d
Beginning with the Fourier integral
Breakup the constant and i xe
1 12 2
i i x f x f e d e d
Call ˆ
f
Fourier transform 1ˆ
2i f f e d f x
F
Inverse Fouriertransform
1 ˆ ˆ
2i x -1 f x f e d f
F
8/13/2019 Lecture 4 fourier series
http://slidepdf.com/reader/full/lecture-4-fourier-series 7/18
1/30/2014
Fourier Transform This function, f (t ), isalmost a sine wave.
Its Fourier transform,peaks at 1000 Hz.
ˆ
f
Function Its Fourier transform
Spectral density and total energy
The Fourier transform, , is also called the spectral density or spectrumbecause it measures the intensity of f (t ) in the frequency interval betweenω and ω + Δ ω .
ˆ
f
Δω
The total signal energy is given by and 2ˆ
E f d 2ˆ
f d
represents the signal energy in the band between ω and ω + Δ ω .
8/13/2019 Lecture 4 fourier series
http://slidepdf.com/reader/full/lecture-4-fourier-series 8/18
1/30/2014
Linearity of the Fourier transform
The Fourier transform is linear.
2 23
5 / 202 2
1 1and
3 2 3 10 xe
e e x
F F
Example – Suppose one wishes to find the Fourier transform of25
2 2
210
3 xe
x
and one notes on page 536 of the textbook that
af x bg x a f x b g x F F F
Then
2 25 5
2 2 2 2
2 1
10 2 103 3 x x
e e x x
F F F
2 23 3
/ 20 / 2012 10
2 3 310e e
e e
Fourier transform of derivatives
Let f ( x) be continuous and f ( x) → 0 as | x| → . Further let ' f x dx
Then 2
'
''
f x i f x
f x f x
F F
F F
To apply, let’s take a detour and define a Dirac delta function .
It is not a real function, but a generalization of a function, called ageneralized function.Informally, the Dirac delta function, δ ( x) is a function which is zeroeverywhere except at x = 0 where it is .
The function, δ ( x-a ) is zero everywhere except at x = a where it is .
Additionally, it has the properties that
1 and x a dx f x x a dx f a
8/13/2019 Lecture 4 fourier series
http://slidepdf.com/reader/full/lecture-4-fourier-series 9/18
1/30/2014
Example Solve the following differential equation using Fourier transforms.
'' 9 cos5 f x f x x
'' 9 cos5 f x f x x F F Fourier transform-both sides
'' 9 cos5 f x f x x F F F Linearity
2 9 cos5 f x f x x - F F F Fourier of 2 nd derivative
2 9 5 52
f x f x x x
- F F Fourier of cosine
2 9 5 52
f x f x x x
- F F
2 9 5 52
f x x x
- F Factor
2 2
5 5
2 9 9
x x f x
F - -
Divide
-12 2
5 5
2 9 9
x x f x
F
- - Inverse Fourier
Example '' 9 cos5 f x f x x
8/13/2019 Lecture 4 fourier series
http://slidepdf.com/reader/full/lecture-4-fourier-series 10/18
1/30/2014
Example
-1
2 2
5 5
2 9 9
x x f x
F - -
2 2
5 51 12 29 9
i x i x x x f x e d e d
- -
2 2
5 512 9 92
i x x x f x e d
- -
Definition
Simplify
5 51 1 1 1
2 34 2 34
i x i x f x e e - -
5 51 1 1
34 2 2
i x i xe e -
1cos5
34 x
-
'' 9 cos5 f x f x x
1 1cos
2 2 2
i ii i e e
e e
Convolution
The convolution of functions f and g is defined by f g
f g f p g x p dp f x p g p dp
The Convolution Theorem – Suppose f ( x) and g ( x) are piecewisecontinuous, bounded and absolutely integrable on the x-axis. Then
2 f g f g F F F
8/13/2019 Lecture 4 fourier series
http://slidepdf.com/reader/full/lecture-4-fourier-series 11/18
1/30/2014
If one of the functions has one peakat x 0, the convolution will shift theother function by x 0 and then blurthe outline by an amount thatdepends on the width of the peak.The amplitude will also be affected
by the value of the peak.
Convolution
Signal Processing Input-Output
8/13/2019 Lecture 4 fourier series
http://slidepdf.com/reader/full/lecture-4-fourier-series 12/18
1/30/2014
Discrete Fourier Transform
2π
( x0, f ( x0))
( x1, f ( x1))N equally spaced points
Find a function of the form 1
0
N inx
nn
f x c e that fits all the points
1
00 0
0
N inx
nn
f f x c e
1
11 10
N
inxnn
f f x c e
1
( 1)1 1
0
N inx N
N N nn
f f x c e
N equations and
N unknowns 0 1 2 1, , , , N c c c c
Solve for cn 1
( )
0
1 N inx k
n k k
c f e N
Discrete Fourier Transform
1( )
0
1 N inx k
n k k
c f e N
Define1
( )
0
ˆ N
inx k n n k
k
f Nc f e
0 1n N
Since the points are equally spaced2
k x k N
21
0
ˆi N nk
N n k
k
f f e
Let2 i N w e
1
0
ˆ N
nk n k
k
f f w
1
0
N inx
nn
f x c eCompare to
8/13/2019 Lecture 4 fourier series
http://slidepdf.com/reader/full/lecture-4-fourier-series 13/18
1/30/2014
Discrete Fourier Transform 2 i N w e
1
0
ˆ N
nk n k
k
f f w 0 1n N
To reconstruct the original signal, we find the inverse of F N
0
1
1
ˆ
ˆ
ˆ
ˆ
N
f
f f
f
0
1
1 N
f
f f
f
0 0 0 0
0 1 2 1
0 2 4 2( 1)
0 1 1 2 1 1 1
N
N N
N N N N
w w w w
w w w w F w w w w
w w w w
andˆ
N f F f
We can write this as a matrix equation by letting
1ˆ
N f F f
Discrete Fourier Transform
Problems with this technique.
To be meaningful, one needs many sample points. Imagine a 1000 by1000 matrix.
Unwieldy, computationally intensive. N 2 operations
Need a less computational intensive technique.
8/13/2019 Lecture 4 fourier series
http://slidepdf.com/reader/full/lecture-4-fourier-series 14/18
1/30/2014
Fast Fourier Transform Presented in a paper by J.W. Cooley and J.Tukey, An algorithm for machine calculation of
complex Fourier Series, 1965.
J. Tukey1915-2000
J. W. Cooleyb. 1926
Uses the results of the Discrete FourierTransform, but in a divide and conquer way.One obtains the same results but with N log N operations
To use this method, one must start with 2 points. p
The vector is broken into an even piece and an odd piece.n f
,0 ,2 ,4 , 2 / 2
,1 ,3 ,5 , 1 / 2
ˆ ˆ ˆ ˆ ˆ
, , , ,ˆ ˆ ˆ ˆ ˆ
, , , ,
even ev ev ev ev N N even
odd od od od od N N odd
f f f f f F f
f f f f f F f
,0 ,2 ,4 , 2 ,1 ,3 ,5 , 1, , , , , , , ,even ev ev ev ev N odd od od od od N f f f f f f f f f f
The Fourier transform is then written.
Fast Fourier Transform
,0 ,2 ,4 , 2 ,1 ,3 ,5 , 1, , , , , , , ,even ev ev ev ev N odd od od od od N f f f f f f f f f f
,0 ,2 ,4 , 2 / 2
,1 ,3 ,5 , 1 / 2
ˆ ˆ ˆ ˆ ˆ
, , , ,ˆ ˆ ˆ ˆ ˆ
, , , ,
even ev ev ev ev N N even
odd od od od od N N odd
f f f f f F f
f f f f f F f
One then obtains using the formulasˆ
f
, ,
, ,
ˆ ˆ ˆ
1, 2, , / 2 1ˆ ˆ ˆ
1, 2, , / 2 1
nn ev n od n
nn M ev n od n
f f f n N
f f f n N
8/13/2019 Lecture 4 fourier series
http://slidepdf.com/reader/full/lecture-4-fourier-series 15/18
1/30/2014
Fast Fourier Transform One can continue cutting the vector in half until one is left with
just 2 by 2 matrices.
This process greatly reduces the number of computations
needed.
MATH 5311 – Advanced Engineering Math
Partial Differential EquationsBasic Concepts
Section 12.1
8/13/2019 Lecture 4 fourier series
http://slidepdf.com/reader/full/lecture-4-fourier-series 16/18
1/30/2014
Partial Derivatives
Definition - Consider a function f of three variables, f ( x, y, z )If y and z are held constant and only x is allowed to vary, the partial
derivative of f with respect to x is denoted by or and is defined
by the limit
f x
x f
0
, , , , , ,limh
f x y z f x h y z f x y z
x h
Similar definitions can be give for the partial derivatives of f withrespect to y and to z .
Example : Given 2 2, , 2 cos( ) f x y z xy x yz xyz xyz
2 2, ,2 cos( )
f x y z xy x yz xyz xyz
x x x x
2 22 cos( ) y x yz x yz x xyz x x x
22 2 cos( ) cos( ) y yz x yz x xyz xyz x x x
22 2 sin( ) cos( ) y xyz yz x xyz xyz xyz x
Partial Derivatives
22 2 sin( ) cos( ) y xyz yz xyz xyz xyz
2 3 22 2 sin( ) 2 cos( ) y xyz xy z xyz yz xyz
8/13/2019 Lecture 4 fourier series
http://slidepdf.com/reader/full/lecture-4-fourier-series 17/18
1/30/2014
Partial Differential Equations
Definition – A partial differential equation (PDE) is an equation containing
partial derivatives of the dependent variable.
Example – In each of the following equations, u is the dependent variableand x, y and t are independent variables.
2 2
2 2
2 2 22
2 2
22
2
, ,0
, ,,
, , ,, 2 3 4
, , ,0
x
u x t u x t c
t xu x y u x y
f x y x y
u x y u x y u x y x y x e x x y y
u x y u x y u x y
x x y
Partial Differential Equations
We want to solve a PDE in a certain domain.
Definition – A domain is an open, connected set of points (usually in ( x,y),( x,y,z ) or ( x,y,z,t ))
Definition – An open sent of points is one that does not include itsboundary.
Open set Not 0pen set
(Closed set)Not 0pen set
8/13/2019 Lecture 4 fourier series
http://slidepdf.com/reader/full/lecture-4-fourier-series 18/18
1/30/2014
Partial Differential Equations Definition – A domain is an open, connected set of points (usually in ( x,y),( x,y,z ) or ( x,y,z,t ))
Definition – A connected set is one where any two points can be joined bya path without leaving the set.
Connected set Not connected set
Partial Differential Equations
Definition – A domain is an open, connected set of points (usually in ( x,y),( x,y,z ) or ( x,y,z,t ))
Domain
Not domain, not connected
Not domain, not open
Not domain, not connectedand not open